Rational and Irrational Numbers — Multiple Choice Questions

Choose the correct statement :

1. Reciprocal of every rational number is a rational number.
2. The square roots of all positive integers are irrational numbers.
3. The product of a rational and an irrational number is an irrational number.
4. The difference of a rational number and an irrational number is an irrational number.

Answer

The difference of a rational number and an irrational number is an irrational number.

For example, 2 is rational number, $\sqrt{3}$ is irrational number and 2 - $\sqrt{3}$ is an irrational number.

∴ Option 4, is the correct option.

Every rational number is

1. a natural number
2. an integer
3. a real number
4. a whole number

Answer

Every rational number is a real number.

∴ Option 3, is the correct option.

Between two rational numbers

1. there is no rational number
2. there is exactly one rational number
3. there are infinitely many rational numbers
4. there are only rational numbers and no irrational numbers.

Answer

Between two rational numbers there are infinitely many rational numbers.

∴ Option 3, is the correct option.

Decimal representation of a rational number cannot be

1. terminating
2. non-terminating
3. non-terminating repeating
4. non-terminating non-repeating

Answer

Decimal representation of a rational number is terminating or non-terminating repeating but not non-terminating non-repeating.

∴ Option 4, is the correct option.

The product of any two irrational numbers is

1. always an irrational number
2. always a rational number
3. always an integer
4. sometimes rational, sometimes irrational

Answer

The product of any two irrational numbers is sometimes rational, sometimes irrational

For example, $2\sqrt{3}$ and $3\sqrt{3}$ are two irrational number .

Their product $2\sqrt{3}$ × $3\sqrt{3}$ = 6 × $(\sqrt{3})^2$ = 6 × 3 = 18 which is rational number.

Again, let $2\sqrt{2}$ and $3\sqrt{3}$ be two irrational numbers.

Their product $2\sqrt{2}$ × $3\sqrt{3}$ = 6 × $\sqrt{2}$ × $\sqrt{3}$ = $6\sqrt{6}$ which is an irrational number.

∴ Option 4, is the correct option.

The division of two irrational numbers is

1. a rational number
2. an irrational number
3. either a rational number or an irrational number
4. neither rational number nor irrational number

Answer

The division of two irrational numbers is either a rational number or an irrational number.

For example, let $2\sqrt{3}$ and $3\sqrt{3}$ be two irrational numbers.

$\dfrac{2\sqrt{3}}{3\sqrt{3}}$ = $\dfrac{2}{3}$ is rational number.

Let $2\sqrt{3}$ and $3\sqrt{5}$ be another two irrational numbers.

$\dfrac{2\sqrt{3}}{3\sqrt{5}}$ is an irrational number.

∴ Option 3, is the correct option.

Which of the following is an irrational number

1. $\sqrt{\dfrac{4}{9}}$

2. $\dfrac{\sqrt{12}}{\sqrt{3}}$

3. $\sqrt{7}$

4. $\sqrt{81}$

Answer

$\sqrt{\dfrac{4}{9}}$ = $\dfrac{\sqrt{4}}{\sqrt{9}}$ = $\dfrac{(\sqrt{2})^2}{(\sqrt{3})^2}$ = $\dfrac{2}{3}$ is a rational number .

$\dfrac{\sqrt{12}}{\sqrt{3}}$ = $\dfrac{\sqrt{2 × 2 × 3}}{\sqrt{3}}$ = $\dfrac{2\sqrt{3}}{\sqrt{3}}$ = 2 is a rational number.

$\sqrt{7}$ is an irrational number.

$\sqrt{81}$ = $\sqrt{9 × 9}$ = 9 is a rational number.

∴ Option 3, is the correct option.

Which of the following numbers has terminating decimal representation ?

1. $\dfrac{3}{7}$

2. $\dfrac{3}{5}$

3. $\dfrac{1}{3}$

4. $\dfrac{3}{11}$

Answer

$\dfrac{3}{7}$ = 0.428571429.. non-terminating decimal representation.

$\dfrac{3}{5}$ = 0.6 is terminating decimal representation.

$\dfrac{1}{3}$ = 0.33333... non-terminating decimal representation.

$\dfrac{3}{11}$ = 0.272727273.. non-terminating decimal representation.

∴ Option 2, is the correct option.

Which of the following is an irrational number ?

1. 0.14
2. $0.14\overline{16}$
3. $0.\overline{1416}$
4. 0.4014001400014...

Answer

0.14 is terminating decimal number hence, rational number.

$0.14\overline{16}$ = 0.14161616... is non-terminating repeating decimal number hence, rational number.

$0.\overline{1416}$ = 0.14161416... is non-terminating repeating decimal number hence, rational number.

0.4014001400014... is non-terminating, non-repeating decimal number hence, it is an irrational number.

∴ Option 4, is the correct option.

Which of the following numbers has non-terminating repeating decimal expansion ?

1. $\dfrac{11}{80}$

2. $\dfrac{17}{160}$

3. $\dfrac{63}{240}$

4. $\dfrac{93}{420}$

Answer

$\dfrac{11}{80}$ = 0.1375 has terminating decimal expansion.

$\dfrac{17}{160}$ = 0.10625 has terminating decimal expansion.

$\dfrac{63}{240}$ = 0.2625 has terminating decimal expansion.

$\dfrac{93}{420}$ = 0.221428571..has non-terminating repeating decimal expansion.

∴ Option 4, is the correct option.

A rational number between $\sqrt{2}$ and $\sqrt{3}$ is

1. $\dfrac{\sqrt{2} + \sqrt{3}}{2}$

2. $\dfrac{\sqrt{2} × \sqrt{3}}{2}$

3. 1.5

4. 1.8

Answer

Consider the squares of $\sqrt{2}$ and $\sqrt{3}$

${(\sqrt2)^2}$ = 2 and ${(\sqrt3)^2}$ = 3

Take any rational number between 2 and 3 which is a perfect square of a rational number,

One such number is 2.25 and

2.25 = $(1.5)^2$

$\sqrt{2.25}$ = 1.5

As, $2 \lt 2.25 \lt 3$, it follows that

$\sqrt{2} \lt \sqrt{2.25} \lt \sqrt{3}$

$\sqrt{2} \lt 1.5 \lt \sqrt{3}$

Hence, one rational number between $\sqrt{2}$ and $\sqrt{3}$ is 1.5 .

∴ Option 3, is the correct option.

The decimal expansion of 2 - $\sqrt{3}$ is

1. terminating and non-repeating
2. terminating and repeating
3. non-terminating and non-repeating
4. non-terminating and repeating

Answer

The decimal expansion of 2 - $\sqrt{3}$ is non-terminating, non-repeating as $\sqrt{3}$ is an irrational number, 2 - $\sqrt{3}$ is also an irrational number and decimal expansion of an irrational number is non-terminating, non-repeating.

∴ Option 3, is the correct option.

The decimal expansion of the rational number $\dfrac{33}{2^2 × 5}$ will terminate after

1. one decimal place
2. two decimal places
3. three decimal places
4. four decimal places

Answer

$\dfrac{33}{2^2 × 5}$ = $\dfrac{33}{4 × 5}$ = $\dfrac{33}{20}$ = 1.65

Hence, decimal expansion of the rational number $\dfrac{33}{2^2 × 5}$ will terminate after two decimal place.

∴ Option 2, is the correct option.

$\sqrt{10} × \sqrt{15}$ is equal to

1. $6\sqrt{5}$
2. $5\sqrt{6}$
3. $\sqrt{25}$
4. $10\sqrt{5}$

Answer

$\sqrt{10} × \sqrt{15}$ = $\sqrt{2 × 5} × \sqrt{3 × 5}$ = $\sqrt{2 × 5 × 3 × 5}$ = $5\sqrt{6}$

∴ Option 2, is the correct option.

$2\sqrt{3} + \sqrt{3}$ is equal to

1. $2\sqrt{6}$
2. 6
3. $3\sqrt{3}$
4. $4\sqrt{6}$

Answer

$2\sqrt{3} + \sqrt{3}$ = $\sqrt{3}(2 + 1)$ = $3\sqrt{3}$

∴ Option 3, is the correct option.

The value of $\sqrt{8} + \sqrt{18}$

1. $\sqrt{26}$
2. $2(\sqrt{2} + \sqrt{3})$
3. $5\sqrt{2}$
4. $6\sqrt{2}$

Answer

$\sqrt{8} + \sqrt{18}$ = $\sqrt{2 × 2 × 2} + \sqrt{2 × 3 × 3}$ = $2\sqrt{2} + 3\sqrt{2}$ = $(2 + 3)\sqrt{2}$ = $5\sqrt{2}$

∴ Option 3, is the correct option.

The number $(2 - \sqrt{3})^2$ is

1. a natural number
2. an integer
3. a rational number
4. an irrational number

Answer

$(2 - \sqrt{3})^2$ = $(2)^2 + (\sqrt{3})^2 - 2 × 2 × \sqrt{3}$ = 4 + 3 - $4\sqrt{3}$ = 7 - $4\sqrt{3}$

Since, 7 - $4\sqrt{3}$ is an irrational number therefore, $(2 - \sqrt{3})^2$ is also an irrational number.

∴ Option 4, is the correct option.

If x is a positive rational number which is not a perfect square, then $-5\sqrt{x}$ is

1. a negative integer
2. an integer
3. a rational number
4. an irrational number

Answer

x is a positive rational number and x is not a perfect square.

Then, $\sqrt{x}$ is an irrational number,

Therefore , $-5\sqrt{x}$ is also an irrational number.

∴ Option 4, is the correct option.

If x, y are both positive rational numbers, then $(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})$ is

1. a rational number
2. an irrational number
3. neither rational nor irrational number
4. both rational as well as irrational number

Answer

$(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) = \sqrt{x} × \sqrt{x} - \sqrt{x} × \sqrt{y} - \sqrt{x} × \sqrt{y} - \sqrt{y} × \sqrt{y} \\[1.5em] \Rightarrow (\sqrt{x})^2 - \sqrt{xy} + \sqrt{xy} -(\sqrt{y})^2 = x - y \\[1.5em]$

Since, x, y are both positive rational numbers, so the difference of two positive rational numbers is also a rational number .

Therefore, $x - y$ is also a rational number. Hence, $(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})$ is a rational number.

∴ Option 1, is the correct option.

After rationalising the denominator of $\dfrac{7}{3\sqrt3 - 2\sqrt{2}}$ , we get the denominator as

1. 13
2. 19
3. 5
4. 35

Answer

$\dfrac{7}{3\sqrt3 - 2\sqrt{2}}$

Let us rationalise the denominator,

Then,

$\dfrac{7}{3\sqrt3 - 2\sqrt{2}} = \dfrac{7}{3\sqrt{3} - 2\sqrt{2}} × \dfrac{3\sqrt{3} + 2\sqrt{2}}{3\sqrt{3} + 2\sqrt{2}} \\[1.5em] \Rightarrow \dfrac{7({3\sqrt{3} + 2\sqrt{2}})}{(3\sqrt{3})^2 - (2\sqrt{2})^2} \\[1.5em] \Rightarrow \dfrac{7 × 3\sqrt{3} + 14 × \sqrt{2}}{27 - 8} \\[1.5em] \Rightarrow \dfrac{21\sqrt{3}+ 14\sqrt{2}}{19} \\[1.5em]$

∴ Option 2, is the correct option.

The number obtained on rationalising the denominator of ${\dfrac{1}{\sqrt{7} - 2}}$ is

1. ${\dfrac{\sqrt{7} + 2}{3}}$

2. ${\dfrac{\sqrt{7} - 2}{3}}$

3. ${\dfrac{\sqrt{7} + 2}{5}}$

4. ${\dfrac{\sqrt{7} + 2}{45}}$

Answer

Given,

${\dfrac{1}{\sqrt{7} - 2}}$

Let us rationalise the denominator,

Then,

${\dfrac{1}{\sqrt{7} - 2}} = {\dfrac{1}{\sqrt{7} - 2}} × \dfrac{\sqrt{7} + 2}{\sqrt{7} + 2} \\[1.5em] \Rightarrow \dfrac{\sqrt{7} + 2}{(\sqrt{7})^2 - {2}^2} \\[1.5em] \Rightarrow \dfrac{\sqrt{7} + 2}{7 - 4} \\[1.5em] \Rightarrow \dfrac{\sqrt{7} + 2}{3} \\[1.5em]$

∴ Option 1, is the correct option.

The number $0.\overline{25}$ is equal to

1. $\dfrac{65}{99}$

2. $\dfrac{37}{99}$

3. $\dfrac{5}{9}$

4. $\dfrac{25}{99}$

Answer

Let x = $0.\overline{25}$

x = 0.252525.....      .........(1)

Multiplying both side by 100, we get

100x = 25.252525...     .........(2)

Subtracting equation (2) from equation (1), we get :

⇒ 100x - x = 25.252525..... - 0.252525......

⇒ 99x = 25

⇒ x = $\dfrac{25}{99}$.

Hence, option 4 is the correct option.

The value of $1.99\overline{9}$ in the form of $\dfrac{p}{q}$, where p and q are integers and q ≠ 0, is

1. $\dfrac{19}{20}$

2. $\dfrac{1999}{1000}$

3. 2

4. $\dfrac{1}{9}$

Answer

Let x = $1.99\overline{9}$

x = 1.999999.....      .........(1)

Multiplying both side with 10, we get

10x = 19.99999...     .........(2)

Subtracting equation (2) from equation (1), we get

⇒ 10x - x = 19.99999..... - 1.99999......

⇒ 9x = 18

⇒ x = $\dfrac{18}{9}$ = 2.

Hence, option 3 is the correct option.

Consider the following two statements:

Statement 1: 2^m x 3ⁿ = (2 + 3)^{m + n}, where m, n are positive integers.

Statement 2: If a is a rational number, and m, n are integers, then a^m.aⁿ = a^{m + n}

Which of the following is valid?

1. Both the Statements are true.

2. Both the Statements are false.

3. Statement 1 is true, and Statement 2 is false.

4. Statement 1 is false, and Statement 2 is true.

Answer

According to statement 1 :

2^m x 3ⁿ = (2 + 3)^{m + n}, where m, n are positive integers.

This statement does not reflect any general law of exponents.

∴ Statement 1 is false.

According to statement 2 :

If a is a rational number, and m, n are integers, then a^m.aⁿ = a^{m + n}

This is one of the fundamental laws of exponents. It holds for any rational base, and all integer exponents.

∴ Statement 2 is true.

Hence, option 4 is the correct option.